Question on invertible matrices $(E-A)$

Question

Let $A$ be a $n^{th}$ order square and skew-symmetric matrix, if $(E-A)$ is an invertible matrix show that $(E+A)(E-A)^{-1}$ is an invertible matrix (where $E$ is an identity matrix)

This is a part of a question I have proved that $||(E-A)u|| \geq ||u||$ for any column vector $u \in \mathbb{R^n}$

If we can show that $(E+A)$ is invertible then the proof will be trivial. Any ideas would be appreciated.

Answer 1

Hint: $\det(E+A)=0$ if and only if $A$ has an eigenvector with eigenvalue $-1$. Assuming that for some $v\in \mathbb{R}^n\backslash\{0\}$ $$Av=-v,$$ try to find a contadiction with the assumption that $E-A$ is invertible. (Alternatively, in the same way you'll get a contradiction to the part of the question you already proved).

Answer 2

I am writing an answer to my own question please let me know if I am wrong

1. \begin{align*} ||(E-A)u|| &= \sqrt{((E-A)u)^T(E-A)u}\\ &= \sqrt{u^T (E-A)^T (E-A)u}\\ &= \sqrt{u^T(E^T- A^T)(E-A)u}\\ &= \sqrt{u^T (E^TE - E^TA -A^TE + A^TA)u} \qquad \qquad \text{since}\,\, A^T = -A\\ &= \sqrt{u^T (E - EA + EA + A^TA)u}\\ &= \sqrt{u^T (E + A^TA)u}\\ &= \sqrt{u^TE u+ u^TA^TAu}\\ &= \sqrt{u^Tu + (Au)^TAu}\\ &= \sqrt{||u||^2 + ||Au||^2}\\ ||(E-A)u|| &\geq ||u|| \end{align*}

2. If $E-A$ is invertible $(E-A)^T$ is invertible as well (it is a property of inverse of a matrix)\begin{align*} (E-A)^T &= (E^T-A^T)\\ &= (E-A^T) \qquad\qquad \text{since $A$ is a skew-symmetric matrix}\\ &= (E+A) \end{align*}

Now we can say $(E+A)$ is an invertible matrix. Since both $(E-A)$ and $(E+A)$ are invertible

\begin{align*} (E+A)(E-A)^{-1}\big[(E+A)(E-A)^{-1}\big]^{-1} &= (E+A)(E-A)^{-1} (E-A)(E+A)^{-1} \\ &= (E+A)E(E+A)^{-1}\\ &= E \end{align*}